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2 years ago

Describe an electron cloud. Then, name and explain a particular atomic model that used the concept of the electron cloud.

Chemistry

1 answer:

pochemuha2 years ago

3 0

Electron cloud is the region around the nucleus in an atom where we can locate an electron.

The concept of electron cloud model was introduced by the Schrodinger and Heisenberg. According to this model, it would be difficult to know the position of the electrons in an atom and they are not particles that orbit around the nucleus. We can only expect the electrons to be present in specific areas called the electron clouds around the nucleus. It is the quantum mechanical model that used the concept of electron clouds. According to the model, the electron cloud or an orbital is a space around the nucleus in an atom where the probability of finding an electron is 90%. It explains that electrons show wave nature. It is difficult to determine the exact position and momentum of an electron in an atom.

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In the reaction of nitrogen gas with oxygen gas to produce nitrogen oxide, what is the effect of adding more oxygen gas to the i

Alborosie

<h3>Answer:</h3>

The Equilibrium would shift to produce more NO

<h3>Explanation:</h3>

The reaction is;

N₂(g) + O₂(g) ⇆ 2NO(g)

When a reaction is at equilibrium then the forward reaction rate will be equivalent to the reverse reaction rate. Additionally, the concentration of the reactants and products are the same.
From Le Chatelier's principle, additional reactants favor the formation of more products while additional products favor the formation of more reactants.
For example, when more oxygen is added then more Nitrogen (II) oxide will be formed.
Oxygen is a reactant and when increased it favors forward reaction which leads to the formation of more NO which is the product.

3 0

2 years ago

James was given a sample of sodium bicarbonate NaHCO3 in a weighing dish to analyze. His teacher asked him to find the percent o

gtnhenbr [62]

This question could be answered easily if the results of the abundance of the other elements are given. You will just have to subtract the sum of all their abundances to 100. Since it's not given, the solution would just be:

Na = 23 g/mol* 1 = 23 g
H = 1 g/mol * 1 = 1 g
C = 12 g/mol * 1 = 12 g
O = 16 g/mol * 3 = 48 g
Total = 84 g

% O = 48/84 * 100 = <em>57.14%</em>

5 0

2 years ago

Read 2 more answers

If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products

puteri [66]

<u>Answer:</u>

<u>For a:</u> The equilibrium mixture contains primarily reactants.

<u>For b:</u> The equilibrium mixture contains primarily products.

<u>Explanation:</u>

There are 3 conditions:

When $K_{eq}>1$ ; the reaction is product favored.
When $K_{eq}$ ; the reaction is reactant favored.
When $K_{e}=1$ ; the reaction is in equilibrium.

For the given chemical reactions:

<u>For a:</u>

The chemical equation follows:

$HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}$

As, $K_{eq}$ , the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

<u>For b:</u>

The chemical equation follows:

$H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}$

As, $K_{eq}>1$ , the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

4 0

2 years ago

Water at 4.4°C is to flow through a horizontal, commercial steel pipe having a length of 305 m at the rate of 150 gal/min. If a

Pavel [41]

Answer:

d = 70.5 mm

Explanation:

given,

length of pipe = 305 m

discharge rate = 150 gal/min

pipe diameter = ?

1 gal/min = 6.30902 × 10⁻⁵ m³/s

150 gal/min = 150 × 6.30902 × 10⁻⁵ m³/s

= 9.46 × 10⁻³ m³/s

$h = \dfrac{flv^2}{2gd}$

Q = A V

$h = \dfrac{fl(\dfrac{Q}{A})^2}{2gd}$

$h = \dfrac{fl(\dfrac{Q}{\dfrac{\pi}{4}d^2})^2}{2gd}$

$h= \dfrac{8flQ^2}{\pi^2gd^5}$

f = 0.048 from moody chart using P/D = 0.00015

$\dfrac{1}{d^5}= \dfrac{6.1\times \pi^2\times 9.8}{8\times 0.048\times 305\times 0.00946^2}$

d = 70.5 mm

Diameter of the pipe is equal to 70.5 mm

7 0

2 years ago

Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l

Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

$2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

We are given:

$\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol$

Putting values in above equation, we get:

$-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]$

$\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol$

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

4 0

2 years ago